[LintCode] #141 Sqrt(x)

class Solution {
    /**
     * @param x: An integer
     * @return: The sqrt of x
     */
    public int sqrt(int x) {
        // write your code here
        if (x == 0 || x == 1) {
            return x;
        }
        
        int start = 1;
        int end = x;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            int q = x / mid;
            if (q == mid) {
                start = mid;
            } else if (q < mid) {
                end = mid;
            } else {
                start = mid;
            }
        }
        
        if ( end * end > 0 && end * end <= x ) { // x=2147483647 --> integer overflow T_T
            return end;
        } else {
            return start;
        }
    }
}

T_T 溢出的那个edge case烦死了！擦！